Two wires of the same size are connected in series. Their resistivities are ho1 and ho2. Find the equivalent resistivity of the combination.

The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area.
Given that the two wires have the same size, let their length be $L$ and their cross-sectional area be $A$.
For the first wire with resistivity $\rho_1$:

$$R_1 = \rho_1 \frac{L}{A}$$

For the second wire with resistivity $\rho_2$:

$$R_2 = \rho_2 \frac{L}{A}$$

When two resistors are connected in series, their equivalent resistance $R_{eq}$ is the sum of their individual resistances:

$$R_{eq} = R_1 + R_2$$

Substitute the expressions for $R_1$ and $R_2$:

$$R_{eq} = \rho_1 \frac{L}{A} + \rho_2 \frac{L}{A}$$ $$R_{eq} = (\rho_1 + \rho_2) \frac{L}{A}$$

When the two wires are connected in series, they form a single composite wire.
The total length of this composite wire is $L_{total} = L + L = 2L$.
The cross-sectional area of the composite wire remains $A$.
Let the equivalent resistivity of this combination be $\rho_{eq}$.
Then, the equivalent resistance can also be written as:

$$R_{eq} = \rho_{eq} \frac{L_{total}}{A}$$ $$R_{eq} = \rho_{eq} \frac{2L}{A}$$

From Step 2 and Step 3, we have:

$$(\rho_1 + \rho_2) \frac{L}{A} = \rho_{eq} \frac{2L}{A}$$

We can cancel out the common term $\frac{L}{A}$ from both sides:

$$\rho_1 + \rho_2 = 2 \rho_{eq}$$

Now, solve for $\rho_{eq}$:

$$\rho_{eq} = \frac{\rho_1 + \rho_2}{2}$$

The equivalent resistivity of the combination is $\frac{\rho_1 + \rho_2}{2}$.

Comparing this result with the given options:
[A] $\sqrt{\rho_1\rho_2}$
[B] $2(\rho_1 + \rho_2)$
[C] $\frac{\rho_1 + \rho_2}{2}$
[D] $\rho_1 + \rho_2$

The calculated equivalent resistivity matches option [C].

The final answer is $\text{C}$.